UNIT G482

Module 3
2.3.1
Series & Parallel Circuits
Candidates should be able to :

KIRCHHOFF’S LAWS
1
LAW 1 (K1)

State Kirchhoff’s second law and appreciate that it is a
consequence of conservation of energy.

Apply Kirchhoff’s first and second laws to circuits.

Select and use the equation for the total resistance of two
or more resistors in series.

Recall and use the equation for the total resistance of two
or more resistors in parallel.
The sum of the currents flowing into any point in a
Circuit is equal to the sum of the currents flowing
Out of the point.
i.e.
Σ Iin = Σ Iout
Greek letter ‘sigma’, which means ‘the sum of all’.
LAW 2 (K2)
Consider the circuit shown opposite.

Solve circuit problems involving series and parallel circuits
with one or more sources of e.m.f.

Explain that all sources of e.m.f have an internal resistance.

Explain the meaning of the term terminal pd.

Select and use the equations :
And
E = I(R + r)
E = V + Ir
I
E
As charge flows through the
battery (of e.m.f. E), electrical
R2
energy is supplied to each coulomb.
R1
The charge then flows through the
VR
VL
resistor of resistance (R1) and
through the filament lamp of
resistance (R2). In each of these
components the electrical energy is converted to heat and heat and
light energy respectively.
Energy supplied per coulomb by
The battery (i.e. the e.m.f.)
=
The sum of the energies converted
per coulomb in each component
(i.e. the sum of the pd’s)
E = VR + VL
And since the current (I) is the same at each point in a SERIES circuit :
E = IR1 + IR2
FXA © 2008
UNIT G482
Module 3
2.3.1
Series & Parallel Circuits
This equation expresses KIRCHHOFF’S SECOND LAW (K2) which
states that :
In any closed loop in a circuit, the sum of the emf’s is
equal to the sum of the pd’s around the loop.

As we have seen from our energy analysis of a simple circuit,
If a unit charge follows a closed path in a circuit :
(2) Use Kirchoff’s Laws to calculate the
currents I1, I2 and I3 in the circuit
shown opposite.


i.e. KIRCHOFF’S SECOND LAW is a consequence
Of the PRINCIPLE OF CONSERVATION OF
ENERGY.

Net emf
E1 - E2
6 - 1.5
4.5
I
=
=
=
=
sum of the pd’s
IR1 + IR2
(I x 15) + (I x 30)
45I
= 4.5 =
45
0.1 A
E1=6V
E2=1.5V
I
R1=15Ω
R2=30Ω
I1
I1
5Ω
I3
C
I2
E
D
4V
Applying Kirchhoff 2 to loop FCDEF
So
20Ω
I3 = 4/20 = 0.2 A
Applying Kirchhoff 2 to loop ABCFA
12 = 20 I3 + 5 I1 = (20 x 0.2) + 5 I1 = 4 + 5 I1
I1

EXAMPLES OF THE USE OF KIRCHHOFF’S LAWS
IN THE SOLUTION OF CIRCUIT PROBLEMS
B
I2
I3 = I1 + I2 ……………….. (1)

12V
A
F
Applying Kirchhoff 1 to point F
4 = 20 I3
Total energy supplied = Total energy dissipated
to the charge.
by the charge.
(1) Use Kirchhoff’s Law 2 to determine
the current (I) in the circuit shown
opposite.
Mark in the currents and label
the closed loops ABCDEF as shown.
2
= 8/5 =
1.6 A
The negative sign tells us that I2 flows
in a direction opposite to that chosen.
Substituting for I1 & I2 in equation (1)
0.2 = 1.6 + I2
So
(3) The circuit diagram opposite shows a
battery charger of emf 5.5 V and
internal resistance 1.0 Ω being used
to recharge a cell of emf 1.2 V and
internal resistance 0.20 Ω.
Use Kirchhoff’s Law 2 to determine
the charging current I.
I2 = 0.2 - 1.6 = - 1.4 A
I
1.2V
5.5V
1.0Ω
0.20Ω
I
Applying Kirchhoff 2 to the loop
5.5 - 1.2 = (I x 1.0) + (I x 0.20)
4.3 = 1.2 I
I = 4.3/1.2 =
3.58 A
FXA © 2008
UNIT G482
Module 3

PRACTICE QUESTIONS (1)
1
The diagram opposite shows a
junction in a circuit.
4
1.5 A
0.5 A
Use Kirhhoff’s First Law
( Σ Iin = Σ Iout) to determine
the size and direction of the
current I.
2
3.2 A
I
Use Kirchhoff’s laws to
determine the current
readings shown on the
ammeters (A1, A2 & A3)
in the circuit shown
opposite.
20Ω
A1
10V
A2
0.2 A
3.7 A
1.8 A
5V
0.7 A
20Ω
A3
Use Kirchhoff’s Second Law to
Calculate :
(a) The pd across resistor R.
3
Series & Parallel Circuits
2.3.1
12V
COMBINATION OF RESISTORS

0.2 A
R
RESISTORS IN SERIES
48Ω
(b) The value of R.
3
Use Kirchhoff’s laws to
determine the size and
direction of the current
at point P in the circuit
shown opposite.
Consider three resistors of resistance
R1, R2 and R3 connected IN SERIES
as shown opposite.
4V
P
10Ω
R1
R2
R3
V1
V2
V3
I

According to Kirchhoff’s Law 1 :
The current (I) is the same in
each resistor.

Since energy is conserved :
Total pd across the combination = The sum of the pds across each resistor
20Ω
10Ω
I
(And since V = IR)
From which :
VT
VT = V1 + V2 + V3
IRT = IR1 + IR2 + IR3
RT = R1 + R2 + R3
For any number of resistors connected IN SERIES,
the TOTAL RESISTANCE (RT) is given by :
12V
RT = R1 + R2 + R3
FXA © 2008
UNIT G482
Module 3
4
Series & Parallel Circuits
2.3.1

If (N) resistors having the same resistance (R) are connected
IN PARALLEL, the TOTAL RESISTANCE (RT) is given by :
RESISTORS IN PARALLEL
Consider three resistors of resistance
R1, R2 and R3 connected IN PARALLEL
as shown opposite.

According to Kirchhoff’s Law 1
I
I1
R1
I2
R2
I3
R3
I = I1 + I2 + I3 ……………..(1)

The pd across each resistor = V

From the definition of resistance :
SPECIAL CASE FOR RESISTORS IN PARALLEL
RT = R
N
I

PRACTICE QUESTIONS (2)
1
Calculate the total resistance of the resistor combinations shown
below :
V
(a)
I = V/R
(b)
2Ω
And applying this to equation (1) :
3Ω
V = V + V + V
RT
R1 R2
R3
From which :
For any number of resistors connected IN PARALLEL, the TOTAL
RESISTANCE (RT) of the combination is given by :
1
RT

2
= 1 + 1 + 1
R1
R2
R3
For resistors connected IN PARALLEL :

The LOWEST value resistors carry the GREATEST proportion
of the current.

6Ω
1 = 1 + 1 + 1
RT
R1 R2 R3
The TOTAL RESISTANCE of the combination is LESS than the
SMALLEST resistance in the combination.
5Ω
10Ω
10Ω
15Ω
20Ω
30Ω
4Ω
(a) What resistor value needs to be connected in parallel with a 20 Ω
resistor to give a combined total resistance of 10 Ω ?
(b) You are provided with several 100 Ω resistors. How could you
combine the minimum number of these resistors to give a total
resistance of 250 Ω ?
3
You are provided with a 400 Ω resistor and two 200 Ω resistors.
Calculate the total resistances which may be obtained by connecting
some or all of these resistors in various combinations.
FXA © 2008
UNIT G482
4
Module 3
Series & Parallel Circuits
2.3.1
For the resistor network shown
calculate :
(a) The total resistance of the
combination.
5A
(b) The pd (V) across each resistor.
I1
4Ω
I2
10Ω
I3
20Ω

EMF (E), TERMINAL PD (V) AND INTERNAL RESISTANCE (r)

All sources of emf have some INTERNAL RESISTANCE (r) , since
they are made from materials (e.g. metal wires, electrodes, chemical electrolytes)
which have some electrical resistance.

If a voltmeter is connected across
the terminals of an electrical supply
(e.g. a cell or battery), it indicates what is
called the TERMINAL PD (V) of the
cell or battery.
5
5A
(c) The value of the currents
(I1, I2 & I3) through each of
the resistors.
cell or battery
5
E
For the circuit shown opposite,
calculate :
If the cell or battery is not part of
an external circuit and the voltmeter
is ‘ideal’ (i.e. it has infinite resistance), then
zero current is drawn and :
240V
(a) The resistance (RAB) of the
parallel combination.
I
(b) The total resistance (RT)
of the circuit.
I1 150Ω
40Ω
A
I2 50Ω
Voltmeter reading = E
voltmeter reading = cell emf (E)
(d) The pd across the 40 Ω
resistor.
The emf (E) of a cell or battery can be
defined as its TERMINAL PD when it is
NOT supplying a current.
In the circuit shown opposite
the voltmeter gives a reading
of 8 V. Calculate :
(a) The pd’s V and V1.
V
400Ω
1600Ω
V1
I
(b) The pd across the 1600 Ω
resistor when the voltmeter
Is disconnected.
V
B
(c) The currents I, I1 and I2.
6
r
V
FXA © 2008
UNIT G482
2.3.1
Module 3
If an external circuit is
connected to the cell or
battery (or the voltmeter
is not perfect and draws
some current), the reading
on the voltmeter drops to
a value less than E.
Series & Parallel Circuits
E
r
V
I

MEASURING EMF (E) AND INTERNAL RESISTANCE (r)

A good estimate of the emf (E) of a cell, battery or power supply
may be obtained by simply measuring the terminal pd with a DIGITAL
VOLTMETER (These have a high resistance and will therefore only
draw a very small current).
I
Voltmeter reading = V (<E)
Value of E given by the digital voltmeter =
R
This is because when there
is a current through the cell,
some of its energy is converted
into heat by the cell’s internal
resistance.

The reading (V) which is < E indicated by the voltmeter is the
TERMINAL PD of the cell and also the pd across the resistor R.
Applying Kirchhoff’s Law 2 to the circuit :
+
pd across the internal resistance
E = V + Ir

E = IR + Ir = I (R + r)
I =
E
(R + r)
V
PROCEDURE
The decrease in voltage is called the ‘LOST VOLTS’ of the cell
and it is proportional to the current.
Emf of the cell = terminal pd
(= pd across R)
6

The circuit shown opposite is
used to obtain a more accurate
determination of the emf (E)
of a cell as well as its internal
resistance (r).
E
r
I
A
V
Corresponding values of the
current (I) in the cell and
pd (V) across the cell are
obtained by adjusting the
variable resistor. The results
are recorded in the table
shown below.
RESULTS

PD (V)/V
Current (I)/A
FXA © 2008
UNIT G482
Module 3

SOME EFFECTS OF INTERNAL RESISTANCE

Low voltage sources from which large currents are drawn (e.g. car
batteries) should have LOW internal resistance, otherwise their
terminal PD (V = E - Ir) would be very low.

High voltage sources (e.g. 5 kV supplies which are sometimes used in
schools) have a HIGH internal resistance so as to limit the current
supplied should they be short-circuited accidentally.

The headlamps on a car will dim if the vehicle is started while they are
switched on. This is because the starter motor draws a large current
and this causes the battery terminal PD (V = E - Ir) to drop sharply.

In order to get an efficient transfer of energy from a source of emf
to an external component of resistance (R), the internal resistance (r)
of the source << R.
E = V + Ir

PRACTICE QUESTIONS (3)
V = -Ir + E
1
A battery of emf 4.5 V and internal resistance 1.5 Ω is connected
to a 10 Ω fixed resistor. Draw a circuit diagram of the arrangement
and calculate the current in the circuit.
2
A battery of emf 15 V and internal resistance 1.2 Ω is connected to
an 8 Ω resistor. Calculate :
GRAPH

Series & Parallel Circuits
2.3.1
When a graph of pd (V) is
plotted, a best fit straight
line should be obtained as
shown opposite.
PD (V)/V
Gradient of line = -r
E
NOTE : In practice the graph is
curved because (r) changes
as the current (I) increases.
Current (I)/A
ANALYSIS OF GRAPH

The equation which links EMF (E), TERMINAL PD (V), CURRENT(I)
and INTERNAL RESISTANCE (r) is :
Rearranging :
Comparing with the equation
For a straight line :
y = mx + c
It can be seen that :
7
(a) The total resistance of the circuit.

Intercept on the y-axis
= E =
V
(b) The current through the battery.

Gradient of the V/I graph = r =
Ω
(c) The “lost volts”
(d) The battery’s terminal pd.
FXA © 2008
UNIT G482
3
4
5
Module 3
2.3.1
Series & Parallel Circuits
A battery of emf (E) and internal resistance (r) was connected in
series with a variable resistor of resistance (R) and an ammeter. If
the ammeter reading was 2.0 A when R was set to 4.0 Ω and it
dropped to 1.5 A when R was set to 6.0 Ω, calculate the values of
E and r.

HOMEWORK QUESTIONS
1
(a) On which conservation
laws are Kirchhoff’s first
and second laws based ?
1.5 V
(b) For the circuit shown
opposite, calculate :
The pd across the terminals of a battery is found to be 3.0 V when
it is measured using a very high resistance voltmeter. The battery
is then connected to a 10 Ω resistor and its terminal pd drops to
2.8 V. Calculate the internal resistance of the battery.
0.09 A
(iii) The resistance of resistor R.
2
(a) State Kirchhoff’s Second
Law.
(b) Apply Kirchhoff’s Second
Law to the circuit shown
opposite to determine
the current at point X.
(c) The value of the resistance R.
A car battery has an emf of 12 V and an internal resistance of
0. 05 Ω. The current drawn by the starter motor is 96 A.
(a) Calculate the terminal pd of the battery when the car is being
started.
(b) If the headlamps are rated at 12 V, 36 W, what is their
resistance ?
(c) Calculate the value of their power output when the starter
motor is in operation.
125 Ω
(ii) The pd across resistor R.
4.5 V
(b) The internal resistance of the cell.
6
R
(i) The pd across the
125 Ω resistor.
A high resistance voltmeter gives a reading of 1.5 V when connected
to a dry cell on “open circuit”. When the cell is connected to a lamp
of resistance R, there is a current of 0.30 A and the voltmeter
reading drops to 1.2 V. Calculate :
(a) The emf of the cell.
8
X
15 Ω
40 Ω
10 V
3
You are given three resistors of resistance 4 Ω, 6 Ω and 8 Ω.
Using all the resistors draw the combination to give :
(a) The largest resistance.
(b) The smallest resistance.
In each case calculate the total resistance of the combination.
FXA © 2008
UNIT G482
4
Module 3
2.3.1
9
Series & Parallel Circuits
6
(a) State Kirchhoff’s First Law.
The diagram below shows a circuit diagram including three resistors.
(b) The diagram below shows part of an electrical circuit.
5.0 V
X
31 mA
I2
22 mA
A
I1
A
20 mA
B
C
30 Ω
I3
(i) Name the component marked X.
(a) The variable resistor is set on its maximum resistance of 20 Ω.
(ii) Determine the magnitude of the currents I1, I2 and I3.
(OCR AS Physics - Module 2822 - January 2004)
5
10 Ω
The results of an experiment to determine the emf (E) and internal
resistance (r) of a power supply are given in the table below.
V/V
1.43
1.33
1.18
1.10
0.98
I/A
0.10
0.30
0.60
0.75
1.00
Calculate the resistance between points :
(i) B and C.
(ii) A and C.
(b) In the circuit shown in the diagram above, the battery has
negligible internal resistance and an emf of 5.0 V. The vaiable
resistor is now set on its lowest resistance of 0 Ω.
Calculate the reading on the ammeter.
(OCR AS Physics - Module 2822 - January 2006)
Plot a suitable graph and use it to find E and r.
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